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19.6 Capacitors in series and parallel  (Page 2/2)

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Capacitors in parallel

[link] (a) shows a parallel connection of three capacitors with a voltage applied. Here the total capacitance is easier to find than in the series case. To find the equivalent total capacitance C p size 12{ {C} rSub { size 8{p} } } {} , we first note that the voltage across each capacitor is V size 12{V} {} , the same as that of the source, since they are connected directly to it through a conductor. (Conductors are equipotentials, and so the voltage across the capacitors is the same as that across the voltage source.) Thus the capacitors have the same charges on them as they would have if connected individually to the voltage source. The total charge Q size 12{Q} {} is the sum of the individual charges:

Q = Q 1 + Q 2 + Q 3 . size 12{Q= {Q} rSub { size 8{1} } + {Q} rSub { size 8{2} } + {Q} rSub { size 8{3} } } {}
Part a of the figure shows three capacitors connected in parallel to each other and to the applied voltage. The total capacitance when they are connected in parallel is simply the sum of the individual capacitances. Part b of the figure shows the larger equivalent plate area of the capacitors connected in parallel, which in turn can hold more charge than the individual capacitors.
(a) Capacitors in parallel. Each is connected directly to the voltage source just as if it were all alone, and so the total capacitance in parallel is just the sum of the individual capacitances. (b) The equivalent capacitor has a larger plate area and can therefore hold more charge than the individual capacitors.

Using the relationship Q = CV size 12{Q= ital "CV"} {} , we see that the total charge is Q = C p V size 12{Q= {C} rSub { size 8{p} } V} {} , and the individual charges are Q 1 = C 1 V size 12{ {Q} rSub { size 8{1} } = {C} rSub { size 8{1} } V} {} , Q 2 = C 2 V size 12{ {Q} rSub { size 8{2} } = {C} rSub { size 8{2} } V} {} , and Q 3 = C 3 V size 12{ {Q} rSub { size 8{3} } = {C} rSub { size 8{3} } V} {} . Entering these into the previous equation gives

C p V = C 1 V + C 2 V + C 3 V . size 12{ {C} rSub { size 8{p} } V= {C} rSub { size 8{1} } V+ {C} rSub { size 8{2} } V+ {C} rSub { size 8{3} } V} {}

Canceling V size 12{V} {} from the equation, we obtain the equation for the total capacitance in parallel C p size 12{C rSub { size 8{p} } } {} :

C p = C 1 + C 2 + C 3 + . . . . size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}

Total capacitance in parallel is simply the sum of the individual capacitances. (Again the “ ... ” indicates the expression is valid for any number of capacitors connected in parallel.) So, for example, if the capacitors in the example above were connected in parallel, their capacitance would be

C p = 1 . 000 µF + 5 . 000 µF + 8 . 000 µF = 14 . 000 µF . size 12{ {C} rSub { size 8{p} } =1 "." "00" µF+5 "." "00" µF+8 "." "00" µF="14" "." 0 µF} {}

The equivalent capacitor for a parallel connection has an effectively larger plate area and, thus, a larger capacitance, as illustrated in [link] (b).

Total capacitance in parallel, C p size 12{C rSub { size 8{p} } } {}

Total capacitance in parallel C p = C 1 + C 2 + C 3 + . . . size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}

More complicated connections of capacitors can sometimes be combinations of series and parallel. (See [link] .) To find the total capacitance of such combinations, we identify series and parallel parts, compute their capacitances, and then find the total.

The first figure has two capacitors, C sub1 and C sub2 in series and the third capacitor C sub 3 is parallel to C sub 1 and C sub 2. The second figure shows C sub S, the equivalent capacitance of C sub 1 and C sub 2, in parallel to C sub 3. The third figure represents the total capacitance of C sub S and C sub 3.
(a) This circuit contains both series and parallel connections of capacitors. See [link] for the calculation of the overall capacitance of the circuit. (b) C 1 size 12{ {C} rSub { size 8{1} } } {} and C 2 size 12{ {C} rSub { size 8{2} } } {} are in series; their equivalent capacitance C S size 12{ {C} rSub { size 8{S} } } {} is less than either of them. (c) Note that C S size 12{ {C} rSub { size 8{S} } } {} is in parallel with C 3 size 12{ {C} rSub { size 8{3} } } {} . The total capacitance is, thus, the sum of C S size 12{ {C} rSub { size 8{S} } } {} and C 3 size 12{ {C} rSub { size 8{3} } } {} .

A mixture of series and parallel capacitance

Find the total capacitance of the combination of capacitors shown in [link] . Assume the capacitances in [link] are known to three decimal places ( C 1 = 1.000 µF , C 2 = 5.000 µF , and C 3 = 8.000 µF ), and round your answer to three decimal places.

Strategy

To find the total capacitance, we first identify which capacitors are in series and which are in parallel. Capacitors C 1 size 12{ {C} rSub { size 8{1} } } {} and C 2 size 12{ {C} rSub { size 8{2} } } {} are in series. Their combination, labeled C S size 12{ {C} rSub { size 8{S} } } {} in the figure, is in parallel with C 3 size 12{ {C} rSub { size 8{3} } } {} .

Solution

Since C 1 size 12{ {C} rSub { size 8{1} } } {} and C 2 size 12{ {C} rSub { size 8{2} } } {} are in series, their total capacitance is given by 1 C S = 1 C 1 + 1 C 2 + 1 C 3 size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } } {} . Entering their values into the equation gives

1 C S = 1 C 1 + 1 C 2 = 1 1 . 000 μF + 1 5 . 000 μF = 1 . 200 μF . size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } = { {1} over {1 "." "000"" μF"} } + { {1} over {5 "." "000"" μF"} } = { {1 "." "200"} over {"μF"} } } {}

Inverting gives

C S = 0 . 833 µF . size 12{ {C} rSub { size 8{S} } =0 "." "833" µF} {}

This equivalent series capacitance is in parallel with the third capacitor; thus, the total is the sum

C tot = C S + C S = 0 . 833 μF + 8 . 000 μF = 8 . 833 μF . alignl { stack { size 12{C rSub { size 8{"tot"} } =C rSub { size 8{S} } +C rSub { size 8{S} } } {} #=0 "." "833"" μF "+ 8 "." "000"" μF" {} # =8 "." "833"" μF" {}} } {}

Discussion

This technique of analyzing the combinations of capacitors piece by piece until a total is obtained can be applied to larger combinations of capacitors.

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Section summary

  • Total capacitance in series 1 C S = 1 C 1 + 1 C 2 + 1 C 3 + . . . size 12{ { {1} over { {C} rSub { size 8{S} } } } = { {1} over { {C} rSub { size 8{1} } } } + { {1} over { {C} rSub { size 8{2} } } } + { {1} over { {C} rSub { size 8{3} } } } + "." "." "." } {}
  • Total capacitance in parallel C p = C 1 + C 2 + C 3 + . . . size 12{ {C} rSub { size 8{p} } = {C} rSub { size 8{1} } + {C} rSub { size 8{2} } + {C} rSub { size 8{3} } + "." "." "." } {}
  • If a circuit contains a combination of capacitors in series and parallel, identify series and parallel parts, compute their capacitances, and then find the total.

Conceptual questions

If you wish to store a large amount of energy in a capacitor bank, would you connect capacitors in series or parallel? Explain.

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Problems&Exercises

Find the total capacitance of the combination of capacitors in [link] .

A circuit is shown with three capacitors. Two capacitors, of ten microfarad and two point five microfarad capacitance, are in parallel to each other, and their combination is in series with a zero point three zero microfarad capacitor.
A combination of series and parallel connections of capacitors.

0.293 μF

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Suppose you want a capacitor bank with a total capacitance of 0.750 F and you possess numerous 1.50 mF capacitors. What is the smallest number you could hook together to achieve your goal, and how would you connect them?

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What total capacitances can you make by connecting a 5 . 00 µF size 12{8 "." "00" mF} {} and an 8 . 00 µF size 12{8 "." "00" mF} {} capacitor together?

3 . 08 µF size 12{3 "." "08" µF } {} in series combination, 13 . 0 µF size 12{"13" "." "0 "µF} {} in parallel combination

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Find the total capacitance of the combination of capacitors shown in [link] .

The circuit includes three capacitors. A zero point three zero microfarad capacitor and a ten microfarad capacitor are connected in series, and together they are connected in parallel with a two point five microfarad capacitor.
A combination of series and parallel connections of capacitors.

2 . 79 µF size 12{2 "." "79"" µF"} {}

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Find the total capacitance of the combination of capacitors shown in [link] .

The figure shows a circuit that is a combination of series and parallel connections of capacitors. On the left of the circuit is a five point zero microfarad capacitor in series with a three point five microfarad capacitor. In the middle is an eight point zero microfarad capacitor. On the right, a zero point seven five microfarad capacitor is in parallel with a fifteen microfarad capacitor, and together they are in series with a one point five microfarad capacitor. Altogether, the system of capacitors on the left, the capacitor in the middle, and the system of capacitors on the right are connected in parallel.
A combination of series and parallel connections of capacitors.
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Unreasonable Results

(a) An 8 . 00 µF size 12{8 "." "00" mF} {} capacitor is connected in parallel to another capacitor, producing a total capacitance of 5 . 00 µF size 12{5 "." "00" mF} {} . What is the capacitance of the second capacitor? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

(a) –3 . 00 µF size 12{8 "." "00" mF} {}

(b) You cannot have a negative value of capacitance.

(c) The assumption that the capacitors were hooked up in parallel, rather than in series, was incorrect. A parallel connection always produces a greater capacitance, while here a smaller capacitance was assumed. This could happen only if the capacitors are connected in series.

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Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
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Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
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SIMRAN Reply
-42m²+60m-18
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-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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